The following assumptions are made for the analysis of heat flow through the fin:

1. Steady state conduction.
2. No heat generation within the fin.
3. Uniform heat transfer coefficient (h) over the entire surface of the fin.
4. Homogeneous and isotropic fin material (i.e. thermal conductivity of material constant).
5. Negligible contact thermal resistance.
6. One-dimensional Heat conduction.
7. Negligible radiation.

Rectangular fin of uniform cross-section

Consider a rectangular fin as shown in fig. of length ‘l’, temperature of the surface over which the fin is attached is 'ts' and the ambient temperature is 'ta'.

Consider an elemental layer at a distance of ‘x’ having thickness ‘dx’. The temperature at the start of elemental layer is ‘t’, While the temperature at a distance ‘x+dx’ is t+dtdxdx.

Heat influx through the elemental area due to conduction

Qx = -K.Acdtdx …(1)

Heat out flux through elemental area due to conduction

Q(x + dx) = -K.Acddx(t + dtdxdx) …(2)

Heat lost from elemental area due to convection

Qconv = h.As(t - ta) …(3)

Balancing heat transfer through fin

Qx = Q(x + dx) + Qconv

-K.Acdtdx = -K.Acdtdx - K.Acd2tdx2dx + h(P.dx)(t - ta)

K.Acd2tdx2dx = h(P.dx)(t - ta)

∴ d2tdx2 = h.PK.Ac(t - ta) assuming, h.PK.Ac = m2

∴ d2tdx2 = m2(t - ta) ...(4)

if, θ = t-ta …(difference of temperature between fin surface and atmosphere)

∴dθdx = dtdx

∴d2θdx2 = d2tdx2

∴ (4) becomes

d2θdx2 - m2θ = 0 …(A)

The above equation is second order homogeneous differential equation and the solution for which is given by

C1emx + C2e - mx = θ …(B)

Where C1 and C2 are constants, can be determined by boundary conditions as below

(1) at x=0, θ0=(ts-ta)

(2) at x=L, dθdx=0 …(∵Q=0 at end, ∴-KAcdtdx=0, ∴dtdx=0)

Substituting condition (1) in (B)

C1+ C2= θ0 …(5)

substituting condition (2) in dθdx

dθdx=mC1emx- C2e-mx

∴0=mC1emL- C2e-mL …(6)

mC1emL= C2e-mL

∴C1=C2 e-mLemL …(7)

∴From (5)

C2e-mLemL+C2=θ0

∴C2=θ0 emLemL+e-mL

Substitute C2 in equation (7)

C1=θ0.e-mLemL+e-mL

Substituting C1 and C2 in (B)

(θ0.e-mLemL+e-mL)emx+(θ0 emLemL+e-mL)e-mx=θ

∴θ=θ0[em(L-x)emL+e-mL+ e-m(L-x)emL+e-mL]

∴θ=θ0[2cosh [m(L-x)] 2cosh (mL) ]

(t-ta)=(ts-ta) cosh [m(L-x)] cosh (mL)

t=ta+(ts-ta) cosh [m(L-x)] cosh (mL) →Temperature distribution equation

Heat transfer through is given by

Q|x=o=-KAcdtdx|x=0

Q=-KAcddt [ta+(ts-ta) cosh [m(L-x)] cosh (mL) ]x=0

Q=-KAc[0+(ts-ta) sinh [m(L-x)] cosh (mL) ×(-m)]x=o

Q=KAchPKAc (ts-ta) tanh⁡(mL)

∴Q=h.P.K.Ac (ts-ta)tanh (mL) →Heat Transfer equation